Question

An airplane originally at rest on a runway accelerates uniformly at 6.0 meters
per second2 for 12 seconds. During this 12-second intervai, the airplane
travels a distance of approximately

Answer 1

432 m

Answer 2

Suvat again, so use the equation s=u^2 + 1/2at^2
u = 0 here so its just 0.5*6*(12)^2 which is 432, just to clarify if that was unclear how he got to that.

Answer 3

a = 6m/s^2
initial velocity = vi = 0
t = 12s

distance = s = (vi)(t) + 0.5(a)(t)^2
= (0)(12) + 0.5(6)(12)^2
= 0 + 432
= 432 m